Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 3 - Determinants - 3.3 Exercises - Page 186: 7

Answer

$ x_{1}=\frac{5s+4}{6(s^{2}-3)}$ $ x_{2}=\frac{-4s-15}{4(s^{2}-3)}$

Work Step by Step

Writing the equation in matrix form $A\textbf{x}=\textbf{b} $ $\begin{bmatrix}6s&4\\9&2s\end{bmatrix} \,\,\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix} =\begin{bmatrix}5\\-2\end{bmatrix}$ To have a unique solution, $\textbf{A}$ must be invertible. This happens if and only if $\textbf{det A} \ne0$ $\textbf{det }\begin{bmatrix}6s&4\\9&2s\end{bmatrix}=12s^{2}-36$ To find the limits on s; $12s^{2}-36=0$ $12s^{2}=36$ $s^{2}=3$ $s=+\sqrt 3, or\,-\sqrt 3$ Therefore for a unique solution, $s\ne+\sqrt 3, or\,s\ne-\sqrt 3$ Compute: $A_{1}\textbf{(b)}=\begin{bmatrix}5&4\\-2&2s\end{bmatrix} ,\, and\,, det(A_{1}\textbf{(b)})=10s+8$ $A_{2}\textbf{(b)}=\begin{bmatrix}6s&5\\9&-2\end{bmatrix} ,\, and\,, det(A_{2}\textbf{(b)})=12s-45$ By using Cramer's rule, if $s\ne+\sqrt 3, or\,s\ne-\sqrt 3$ that is, A is Invertible, then $ x_{1}=\frac{det(A_{1}(\textbf{b}))}{det(A_{})}=\frac{10s+8}{12(s^{2}-3)}=\frac{5s+4}{6(s^{2}-3)}$ $ x_{2}=\frac{det(A_{2}(\textbf{b}))}{det(A_{})}=\frac{-12s-45}{12(s^{2}-3)}=\frac{-4s-15}{4(s^{2}-3)}$
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